Physics force on moving charge problem!?

Question

a 5.0 MeV (kinetic engery) proton enters a 0.20-T field, in a plane perpendicular to the field. What is the radius of its path?

I'm still trying to figure this out. =/
Please provide step-by-step directions. Thanks!

Answer 1

Force = q v B = m v^2 / r (Equation 1)

q v B = Lorentz Force
m v^2 / r = motion in a circle

q is charge
v is velocity
B is mag. field strength
m is mass
v^2 is velocity^2
r is path radius

First we need to find the velocity of the proton....
KE = 5 MeV = 1/2 m v^2
5MeV = 8 x 10^-13 Joules = 1/2 m v^2
8 x 10^-13 = 1/2 x 1.67 x 10^-27 x v^2

v^2 = 8x 10^-13 / (1/2 x 1.67 x 10^-27)
v^2 = 9.581 x 10^14
v = 3.095 x 10^7 ms^-1

now we go back to equation 1...

q v B = m v^2 / r

Re-arrange for r (and cancel for v) gives-:

r = m v / q B

Just put in the numbers....

r = (1.67 x 10^-27) x (3.095 x 10^7) / (1.6 x 10^-19) x 0.20

r = 1.6152 metre radius

Hope you understand how to do this, this is fundamental in the realm of atomic and nuclear physics - which I happen to lecture.